# Path With Maximum Minimum Value in Python

PythonServer Side ProgrammingProgramming

Suppose we have a matrix A of integers with R rows and C columns, we have to find the maximum score of a path starting from [0,0] and ending at [R-1,C-1]. Here the scoring technique will be the minimum value in that path. For example, the value of the path 8 → 4 → 5 → 9 is 4. A path moves some number of times from one visited cell to any neighboring unvisited cell in one of the 4 cardinal directions (north, east, west, south).

For example, if the grid is like −

 5 4 5 1 2 6 7 4 6

The orange cells will be the path. The output is 4

To solve this, we will follow these steps −

• r := number of rows and c := number of columns
• ans := minimum of A[0, 0] and A[r – 1, c – 1]
• make one matrix called visited of the order same as A, and fill this with FALSE
• h := a list, where we store a tuple (-A[0, 0], 0, 0)
• Make heap from h
• while h is not empty
• v, x, y := delete the h from heap, and store three values
• if x = r – 1 and y := c – 1, then come out from loop
• ans := min of ans, A[x, y]
• visited[x, y] := true
• for dy, dx in the list [(-1, 0), (1, 0), (0, 1), (0, -1)], do
• a := x + dx and b := y + dy
• if a in range 0 to r – 1 and b in range 0 to c – 1 and visited[a, b] is false,
• insert (-A[a, b], a, b) into heap with h
• return ans

Let us see the following implementation to get better understanding −

## Example

import heapq
class Solution(object):
def maximumMinimumPath(self, A):
"""
:type A: List[List[int]]
:rtype: int
"""
r,c = len(A),len(A[0])
ans = min(A[0][0],A[-1][-1])
visited = [[False for i in range(c)] for j in range(r)]
h = [(-A[0][0],0,0)]
heapq.heapify(h)
while h:
# print(h)
v,x,y = heapq.heappop(h)
if x== r-1 and y == c-1:
break
ans = min(ans,A[x][y])
visited[x][y]= True
for dx,dy in {(-1,0),(1,0),(0,1),(0,-1)}:
a,b = x+dx,y+dy
if a>=0 and a<r and b>=0 and b<c and not visited[a][b]:
heapq.heappush(h,(-A[a][b],a,b))
return ans

## Input

[[5,4,5],[1,2,6],[7,4,6]]

## Output

4
Published on 05-Mar-2020 07:10:38